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Bunuel
Is 3^(p-2) greater than 1,000?

(1) 3^(p+1)< 54,000 --> \(3^{p+1}< 27*2,000\) --> \(3^{p+1}< 3^3*2,000\) --> divided both sides by 3^3: \(3^{p-2}< 2,000\). Thus 3^(p-2) may or may not be greater than 1,000. Not sufficient.

(2) 3^p < 3^(p-1) + 2,000 --> \(3^p -3^{p-1}< 2,000\) --> factor out 3^(p-2): \(3^{p-2}(3^2 -3)< 2,000\) --> \(3^{p-2}< \frac{1,000}{3}\). Sufficient.

Answer: B.

Bunuel,

Could you please explain why you can factor out 3^p-2 in statement 2?

And, for my info, would it be sufficient to just pick numbers in (1) ?

Edit:

I think I got it. I tested it if p=4 you can factor out 2 and you get your factorization. Is there a common formula to remember how to factor out expontents this way? Since I now myself I will have forgotten this in a week from now...
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Bunuel
Is 3^(p-2) greater than 1,000?

(1) 3^(p+1)< 54,000 --> \(3^{p+1}< 27*2,000\) --> \(3^{p+1}< 3^3*2,000\) --> divided both sides by 3^3: \(3^{p-2}< 2,000\). Thus 3^(p-2) may or may not be greater than 1,000. Not sufficient.

(2) 3^p < 3^(p-1) + 2,000 --> \(3^p -3^{p-1}< 2,000\) --> factor out 3^(p-2): \(3^{p-2}(3^2 -3)< 2,000\) --> \(3^{p-2}< \frac{1,000}{3}\). Sufficient.

Answer: B.

Bunuel,

Could you please explain why you can factor out 3^p-2 in statement 2?

And, for my info, would it be sufficient to just pick numbers in (1) ?

Edit:

I think I got it. I tested it if p=4 you can factor out 2 and you get your factorization. Is there a common formula to remember how to factor out expontents this way? Since I now myself I will have forgotten this in a week from now...

Check backwards: \(3^{p-2}(3^2 -3)=3^{p-2}*3^2 -3^{p-2}*3=3^{p-2+2} -3^{p-2+1}=3^{p} -3^{p-1}\).

It's all about operations involving the same bases:
Keep the base, add or subtract the exponent (add for multiplication, subtract for division)
\(a^n*a^m=a^{n+m}\)

\(\frac{a^n}{a^m}=a^{n-m}\)

Check the links in my previous post for similar questions.

Theory on Exponents: math-number-theory-88376.html

All DS Exponents questions to practice: search.php?search_id=tag&tag_id=39
All PS Exponents questions to practice: search.php?search_id=tag&tag_id=60

Tough and tricky DS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125967.html
Tough and tricky PS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125956.html

Hope it helps.
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Bunuel, do you think this solution makes sense?

3^(p-2) > 1000?

3^p > 3^2 * 10^3?

p > 8?

We have 3, 9, 27, 81, 243, 729 as powers of 3.

(1) 3^p * 3 < 54,000

3^p < 18,000 = 2 * 3^2 * 10^3

3^p > 3^2 * 10^3. Yes
or
3^p = 0. No

We do not know where in this interval 3^p falls. Not sufficient.

Eliminate A, C, E.

(2) 3^p < 3^ (p-1) + 2000

Plug in.

p = 6 --> 729 < 243 + 2000 . Inequality satisfied. p NOT less than 8
p = 7 --> ~2190 < 729 + 2000. Inequality satisfied. p NOT less than 8
p = 8 --> ~6600 < 2190 + 2000. Inequality does not hold. p MUST be less than 8.

Therefore, Statement 2 is sufficient. The answer is B.
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CCMBA
Bunuel, do you think this solution makes sense?

3^(p-2) > 1000?

3^p > 3^2 * 10^3?

p > 8?

We have 3, 9, 27, 81, 243, 729 as powers of 3.

(1) 3^p * 3 < 54,000

3^p < 18,000 = 2 * 3^2 * 10^3

3^p > 3^2 * 10^3. Yes
or
3^p = 0. No

We do not know where in this interval 3^p falls. Not sufficient.

Eliminate A, C, E.

(2) 3^p < 3^ (p-1) + 2000

Plug in.

p = 6 --> 729 < 243 + 2000 . Inequality satisfied. p NOT less than 8
p = 7 --> ~2190 < 729 + 2000. Inequality satisfied. p NOT less than 8
p = 8 --> ~6600 < 2190 + 2000. Inequality does not hold. p MUST be less than 8.

Therefore, Statement 2 is sufficient. The answer is B.


Refer text highlighted in red
1. In case St1 is not sufficient, You can eliminate A and D and not C and E

I don't see a problem with your method :wink:
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[/quote]


Refer text highlighted in red
1. In case St1 is not sufficient, You can eliminate A and D and not C and E

I don't see a problem with your method :wink:[/quote]

Sorry about that. Thanks, Wounded Tiger!
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se7en14
Is 3^(p-2) greater than 1,000?

(1) 3^(p+1)< 54,000
(2) 3^p < 3^(p-1) + 2,000


I tried brute force here why is that wrong ?

3^1= 3
3^2= 9
3^3= 27
3^4= 81
3^5 =243
3^6=729
3^7= 2187
3^8= 6561
3^9 = 19683
3^10=59049

when question is asking Is 3^(p-2) greater than 1,000?
only possible if P -2>6 or p>8

1)3^(p+1)< 54,000
so p+1 <10 or P <9

hence shouldn't this be sufficient .

(2) 3^p < 3^(p-1) + 2,000
after simplification
3^(P-1)<1000

so p-1<=6
p<=7

sufficient.
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se7en14
Is 3^(p-2) greater than 1,000?

(1) 3^(p+1)< 54,000
(2) 3^p < 3^(p-1) + 2,000


I tried brute force here why is that wrong ?

3^1= 3
3^2= 9
3^3= 27
3^4= 81
3^5 =243
3^6=729
3^7= 2187
3^8= 6561
3^9 = 19683
3^10=59049

when question is asking Is 3^(p-2) greater than 1,000?
only possible if P -2>6 or p>8

1)3^(p+1)< 54,000
so p+1 <10 or P <9

hence shouldn't this be sufficient .

(2) 3^p < 3^(p-1) + 2,000
after simplification
3^(P-1)<1000

so p-1<=6
p<=7

sufficient.

We are not told that p is an integer.

Is 3^(p-2) > 1,000 --> is p > ~8.3.
3^(p+1) < 54,000 --> p < ~8.9.

Hope it's clear.
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Bunuel
qlx
se7en14
Is 3^(p-2) greater than 1,000?

(1) 3^(p+1)< 54,000
(2) 3^p < 3^(p-1) + 2,000


I tried brute force here why is that wrong ?

3^1= 3
3^2= 9
3^3= 27
3^4= 81
3^5 =243
3^6=729
3^7= 2187
3^8= 6561
3^9 = 19683
3^10=59049

when question is asking Is 3^(p-2) greater than 1,000?
only possible if P -2>6 or p>8

1)3^(p+1)< 54,000
so p+1 <10 or P <9

hence shouldn't this be sufficient .

(2) 3^p < 3^(p-1) + 2,000
after simplification
3^(P-1)<1000

so p-1<=6
p<=7

sufficient.

We are not told that p is an integer.

Is 3^(p-2) > 1,000 --> is p > ~8.3.
3^(p+1) < 54,000 --> p < ~8.9.

Hope it's clear.

Thank you, was thinking along similar lines too!
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3^(p-2) < 1000?

This is (3^p)*(3^-2) < 1000, which means we can rephrase the question as:

3^p < 9000?

(1) 3^(p+1) = 3^p*3^1 < 54000
meaning: 3^p < 18000

3^p could be less than 9000 or greater than 9000, INSUFFICIENT

(2) 3^p < 3^(p-1) + 2,000

3^p < 3^p*3^-1 + 2000
3^p < 3^p/3 + 2000
3*3^p < 3^p + 6000
2*3^p < 6000
3^p < 3000

3^p is less than 3000, meaning that it will always be less than 9000 as well, SUFFICIENT

Answer B
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